Transcript Op-Amp Applications

EE 174
Spring 2017
Operational Amplifiers
Contents
• Introduction
• Brief of History
• Fundamentals of Op-Amps
• Basic operation
• Gain
• Offset
• Applications
Introduction
• Operational Amplifier (Op-Amp) name comes from the fact that it was originally used to perform
mathematical operations.
• Op-Amp is an active circuit element which is basic component used to build analog circuits.
• Op-Amp is a low cost integrating circuit consisting of transistors, resistors, diodes and capacitors.
• Op-Amp amplify an input signal produces an output voltage equal to the difference between the two
input terminals multiplied by the gain A.
• Op-Amps are two-port networks in which the output voltage or current is directly proportional to
either input voltage or current. Four different kind of amplifiers exits:
•
•
•
•
Voltage amplifier: Av = Vo / Vi
Current amplifier: Ai = Io / Ii
Transconductance amplifier: Gm = Io / Vi
Transresistance amplifier: Rm = Vo / Ii
•
The op-amp always “wants” both inputs (inverting and non-inverting) to be the same value. If they
are not, the same value, the op amp output will go positive or negative saturation, depending on
which input is higher than the other.
•
Op-Amps are commonly used for both linear and nonlinear applications: Inverting/Non-inverting
Amplifiers, Variable Gains Amplifiers, Summers, Integrators/Differentiators, Filters (High, Low, Band
Pass and Notch Filters), Schmitt trigger, Comparators, A/D converters.
Brief History of Op-Amp
Monolithic IC Op-Amp
Vacuum Tube Op-Amps (1930’s-1940’s) Solid State Discrete Op-Amps
Dual-supply voltage of +300/-300 V
(1960’s)
Output swing +/- 50 volts
Open-loop voltage gain of 15,000 to 20,000,
Slew rate of +/- 12 volts/µsecond
Maximum output current of 1 mA
George Philbrick
Dual-supply voltage of +15/-15 V
Output swing +/- 11 volts
Open-loop voltage gain of 40,000,
Slew rate of +/- 1.5 volts/µsecond
Maximum output current of 2.2 mA
• First created in 1963 μA702
by Fairchild Semiconductor
• μA741 created in 1968,
became widely used due to
its ease of use 8 pin, dual
in-line package (DIP)
• Further advancements
include use of field effects
transistors (FET), greater
precision, faster response,
and smaller packaging
Internal Op-Amp Circuits
Op-Amp Ideal, Equivalent Circuit, Characteristics and Features
Ideal Op-Amp
Op-Amp Equivalent
Circuit
Op-Amp Characteristics
Op-Amp Symbol
741 Op-Amp Features
+V2: Non-inverting input
-V1: Inverting input
+Vs: Positive source PS
-Vs: Negative source PS
Vout: Output voltage
ON: Offset Null
NC: Not Connected
The Ideal Op-Amp Assumptions
vd = v+ - vvo = Avd = A(v+ – v-)
Note: v+ = v2, v- = v1
1) The input impedance Ri is infinite - i.e. no current flows into either input.
2) The output impedance Ro is zero - i.e. the op-amp can drive any load impedance to
any voltage.
3) The open-loop gain (A) is infinite.
4) The bandwidth is infinite.
5) The output voltage is zero when the input voltage difference is zero.
6) The slew rate is infinite
Op-Amp Two Basic Operations
The Inverting Op Amp
The Non-inverting Op Amp
The op-amp always “wants” both inputs (inverting and non-inverting) to
be the same value.
Op-Amp Gain
Open loop gain: This form of gain is measured when
no feedback is applied to the op amp.
Closed loop gain: This form of gain is measured when
the feedback loop is closed and the overall gain of the
circuit is much reduced. It has two forms, signal gain
and noise gain.
The expression for the gain of a closed-loop amplifier
involves the open-loop gain. If G is the actual gain, NG
is the noise gain, and AVOL is the open-loop gain of the
amplifier, then:
 G = NG
0
Since the open-loop gain AVOL is very high,
the closed-loop gain of the circuit is simply
the noise gain, G = NG.
Loop gain: The difference between the open-loop gain and the closed-loop gain is known as the loop gain.
This is useful information because it gives you the amount of negative feedback that can apply to the
amplifier system.
Signal Gain and Noise Gain
Signal gain: This is the gain applied to the input signal, with the feedback loop connected. It can be
inverting or non-inverting. It can even be less than unity for the inverting case. Signal gain is the gain that
we are primarily interested in when designing circuits.
Noise gain: Noise gain is the reciprocal of the attenuation from the output of an op amp (or any feedback
loop) to the input. This is the gain applied to a noise source in series with an op amp input. It is also the gain
applied to an offset voltage. It is the same for either an inverting or non-inverting stage. It is the noise gain that
is used to determine stability.
Summary of Signal Gain and Noise Gain
Note: Voltage sources are grounded
when calculating noise gain in op-amp.
The attenuation is R1/(R1 + R2)
 The noise gain is (R2 + R1)/R1
Op-Amp Gain and Bandwidth
The Voltage Gain (A) of the operational amplifier
can be found using the following formula:
and in Decibels or (dB) is given as:
Gain versus Bandwidth
Applying feedback will
reduce the gain but increase
the bandwidth.
Gain Bandwidth Product (GBP)
Gain-bandwidth (GBW) product is defined as the
op-amp gain multiplied by the bandwidth. The
GBW product can be used to calculate the closedloop gain bandwidth.
GBW = A x BW
where A is in ratio (not in dB)
GBW/closed-loop gain = closed-loop BW
Closed-loop BW = GBW / A
Examples: The GBW @ unity gain (from graph)
 GBW = 1 x 10 MHz = 10 MHz
The GBW @ closed-loop gain A = 100
 GBW = 100 x 100 KHz = 10 MHz
BW = 1 MHz, closed-loop gain?
 A = GBW / BW = 10 MHz / 1MHz = 10
Closed loop gain 70 dB. Closed-loop BW?
Gain: 20 log A = 70 dB  log A = 70/20 = 3.5 or
A = 103.5 = 3162
Closed-loop BW = 10 MHz / 3162 = 3163 Hz
Bode Plot Example
Given a transfer function:
H(s) =
=
100
100
=
𝑠
𝑠+30
30 (30 +1)
100
1
1
= 3.3 𝑠
𝑠
30 ( +1)
( +1)
30
30
Constant gain = 3.3
Gain in dB = 20 log (3.3) ≈ 10 dB
Pole at 30 rad/sec
Phase = 45o at 30 rad/sec
Op-Amp Saturation
The op amp has three distinct regions of operation:
• Linear region: −Vsat_L- < Vo < Vsat_L+
• Positive saturation: Vo > Vsat_L+
• Negative saturation: Vo < −Vsat_LThe output voltage of a practical amplifier cannot exceed certain
threshold value is called saturation. A voltage amplifier behaves
linearly, i.e., Vo/Vi = Av = constant as long as the output voltage remains
below the “saturation” voltage,
−Vsat_L- < Vo < Vsat_L+
Note that the saturation voltage, in general, is not symmetric.
For an amplifier with a given gain, Av, the above range of Vo translate
into a certain range for Vi
− Vsat_L- < Vo < Vsat_L+
− Vsat_L- < Av Vi < Vsat_L+
− Vsat_L- / Av < Vi < Vsat_L+ / Av
Any amplifier will enter its saturation region if Vi is raised above certain
limit. The figure shows how the amplifier output clips when amplifier is
not in the linear region.
Vsat_L+
Vsat_L-
Example: For Av = 105, Vsat_L- = -12V, Vsat_L+ = +15V, find range of input Vi to prevent saturation.
Solution: -12V / 105 < Vi < 15V / 105 or -0.12mV < Vi < 0.15mV
Maximum Output Voltage Swing
The maximum output voltage, VOM±, is defined as “the maximum positive or negative peak output voltage that can be
obtained without wave form clipping when quiescent DC output voltage is zero”. VOM± is limited by the output
impedance of the amplifier, the saturation voltage of the output transistors, and the power supply voltages.
Note that VOM± depends on the output load.
The maximum value that VBQ6 can be is +VCC, therefore
VO ≤ +VCC – VR1 – VBEQ6 – VSATQ6. The minimum value that Vi can be is
–VEE, therefore VO ≥ –VEE + VR2 + VBEQ7 + VSATQ7. This emitter follower
structure cannot drive the output voltage to either rail.
Rail to rail output op amps use a common
emitter (bipolar) or common source (CMOS)
output stage. With these structures, the
output voltage swing is only limited by the
saturation voltage (bipolar) or the on
resistance (CMOS) of the output transistors,
and the load being driven.
Op-Amp slew rate basics
Square Wave: The slew rate (SR)of an op amp is the rate of
change in the output voltage caused by a step change on the input.
It is measured as a voltage change in a given time - typically V / µs.
The slew rate may be very important in many applications.
For the given graph, it takes 2 µs to swing 10 V so SR = 5V/ µs.
Sine Waves: The maximum rate of change for a sine wave occurs at the
zero crossing and may derived as follows:
Given vo = VP sin(2π f t)
Slew Rate (SR) =
V/µs = 2π f VP cos(2π f t) =
max
= 2π f VP
t=0
SR = 2π fmax VP
Example: Slew rate S0 = 1 V/μs; input voltage vin (t) = sin(2π × 105t);
closed loop gain AV =10.
1) Sketch the theoretically correct output and the actual output of the amplifier in the same graph.
2) What is the maximum frequency that will not violating the given slew rate?
Solutions:
1) With the closed-loop voltage gain of 10  vo(t) = 10 sin(2π × 105t) 
= Aω = A x 2π x f = 10 × 2π × 105 = 6.28 V/μs
max
2) fmax = SR / (A x 2π) = 1 x 106 / (10 x 2π) = 15.9 kHz
Op-Amp Offset Parameters
Another practical concern for op-amp performance is voltage offset.
A perfect op-amp would output exactly zero volts with both its inputs
shorted together and grounded. However, most op-amps off the
shelf will drive their outputs to a saturated level, either negative or
positive even if the op-amp in question has zero common-mode gain
(infinite CMRR). This deviation from zero is called offset.
In the example shown above, the output voltage is saturated at a
value of positive 14.7 volts, just a bit less than +V (+15 volts) due to
the positive saturation limit of this particular op-amp.
Unlike common-mode gain, there are usually provisions made by the
manufacturer to trim the offset of a packaged op-amp. Usually, two
extra terminals on the op-amp package are reserved for connecting
an external “trim” potentiometer. These connection points are
labeled offset null and are used in this general way:
741 OPAMP have offset voltage null capability by connecting 10 K ohm pot between pin 1 and pin 5.
By varying the potentiometer, output offset voltage (with inputs grounded) can be reduced to zero volts. For the 741C
the offset voltage adjustment range is ± 15 mV.
Op-Amp Offset Example
Example: Determine the output voltage in each of the following cases for the open
loop differential amplifier shown below.
(a) vin 1 = 5 µVdc, vin 2 = -7 µVdc
(b) vin 1 = 10 mV rms, vin 2= 20 mV rms
Specifications of the OPAMP are given below:
A = 200,000, Ri = 2 MΩ , RO = 75Ω, + VCC = + 15 V, - VEE = - 15 V, and output voltage
swing = ± 14V.
Solution:
(a). The output voltage of an OPAMP is given by
Remember that vo = 2.4 V dc with the assumption that the dc output voltage is zero
when the input signals are zero.
(b). The output voltage equation is valid for both ac and dc input signals. The output
voltage is given by
Thus the theoretical value of output voltage vo = -2000 V rms. However, the OPAMP
saturates at ± 14 V. Therefore, the actual output waveform will be clipped as shown
fig. 5. This non-sinusoidal waveform is unacceptable in amplifier applications.
Op-Amp Applications
Differential Amplifier
If R1 = R2 and R3 = R4
Op-Amp Applications
Find i, if and vo.
i=0A
For the ideal op amp shown below, what should be the
value of resistor Rf to obtain a gain of 5?
Op-Amp Applications
1) Calculate all voltage drops and currents in this circuit, complete with arrows for current direction and polarity
markings for voltage polarity. Then, calculate the overall voltage gain of this amplifier circuit (AV), both as a ratio and
as a figure in units of decibels (dB):
Solutions: Gain = AV = 1.468 = 3.335 dB
2) Given gain RF/RS = 100, GBW product = A0 ω0 = K = 4π × 106 .
Determine the overall 3-dB bandwidth of the cascade amplifier Solutions: The 3-dB bandwidth for each amplifier is:
below:
Thus, the bandwidth of the cascade amplifier is 4π × 104,
the cascade amplifier gain is A1A1= 100 × 100 = 104.
To achieve the same gain with a single-stage amplifier
having the same K, the bandwidth would be
Op-Amp Applications
Problem: Suppose that we need an amplifier with input resistance of 500 kΩ or greater and a voltage gain of -10. The
feedback resistors are to be implemented in integrated form and have values of 10 kΩ or less to conserve chip area.
Choose a suitable circuit configuration and specify the resistance values.
Solution:
Gain of -10  Use inverting Op-Amp where gain
A = - R2/R1 = -10
To attain desired input resistance R1 = 500 kΩ  R2 = 5MΩ.
These values exceed the maximum values allowed.
To attain large input resistance with moderate resistances
for an inverting amplifier, we cascade a voltage follower
with an inverter.
To achieve the desired gain = - 10.
Choose R1 = 1 kΩ  R2 = 10 kΩ.
R2
Op-Amp Applications
For Rs = 10 kΩ, RL = 1 kΩ  VL = 0.1 Vin
 Huge attenuation of the signal source.
If voltage follower (buffer amplifier) is used.
 VL = Vout = Vin
Active Filters, Integrators & Differentiators
Consider the circuit shown. Various circuits can be made with
different choices for Z1 and Z2. Following the inverting amplifier
solution, we find:
Active Filters, Integrators & Differentiators
Active Filters, Integrators & Differentiators
Integrator
The Comparators
• A comparator is a device with two input terminals, inverting and noninverting, and an output that usually
swings from rail to rail.
• A comparator has low offset, high gain, and high common-mode rejection.
• A comparator has a logic output that indicates which of the two inputs is at a higher potential. If its output is
TTL or CMOS compatible (and many comparators are), it is always intended to be at one rail or the other—or
making a rapid transition between the two.
• Comparators are designed to work as open-loop systems, to drive logic circuits, and to work at high speed, even
when overdriven.
The Comparators with and without Hysteresis
Comparator without Hysteresis
This configuration uses a voltage divider (Rx and Ry) to set up the
threshold voltage. The comparator will compare the input signal (Vin) to
the threshold voltage (Vth). The comparator input signal is applied to
the inverting input, so the output will have an inverted polarity.
When the Vin > Vth the output will drive to
the negative supply (GND or logic low).
When Vin < Vth the output will drive to the
positive supply (Vcc = 5V or logic high).
Comparator with Hysteresis
A small change to the comparator circuit can be used to add hysteresis. Hysteresis uses two different threshold
voltages to avoid the multiple transitions introduced in the previous circuit. For this inverting comparator, the input
signal must exceed the upper threshold (VH) to transition low or below the lower threshold (VL) to transition high.
The resistor Rh sets the
hysteresis level.
When the output is at a logic high
(5V), Rh is in parallel with Rx. This
drives more current into Ry, raising
the threshold voltage (VH) to 2.7V.
The input signal will have to drive
above VH=2.7V to cause the output
to transition to logic low (0V).
When the output is at logic low (0V),
Rh is in parallel with Ry. This reduces
the current into Ry, reducing the
threshold voltage to 2.3V. The input
signal will have to drive below VL=2.3V
to cause the output to transition to
logic high (5V).
Design of Hysteresis Comparator
The design requirements are as follows:
• Supply Voltage: +5 V
• Input: 0V to 5V
• VL (Lower Threshold) = 2.3V
• VH (Upper Threshold) = 2.7V
• VH - VL = 0.4V
• Vth ± 0.2V = 2.5V ± 0.2V
Equations (1) and (2) can be used to select the
resistors needed to set the hysteresis threshold
voltages VH and VL.
One value (Rx) must be arbitrarily selected. In this
example, Rx was set to 100kΩ to minimize current
draw.
Rh was calculated to be 575kΩ, so the closest
standard value 576kΩ was used
Op-amp Comparator Circuit
Although op amps are not designed to be used as
comparators, there are, nevertheless, many applications
where the use of an op amp as a comparator is a proper
engineering decision.
WHY USE AN OP AMP AS A COMPARATOR?
• Convenience
• Economy
• Low IB
• Low VOS
There are several reasons to use op amps as
comparators. Some are technical, one is purely
economic. It is economical to use the spare op amp in a
quad as a comparator rather than buying an additional
comparator, but this is not a good design practice.
WHY NOT USE AN OP AMP AS A COMPARATOR?
• Speed
• Inconvenient input structures
There are several reasons not to use op amps as comparators. First
• Inconvenient logic structures
and foremost is speed, but there are also output levels, stability (and
• Stability/hysteresis
hysteresis), and a number of input structure considerations.
The Op-Amp Comparators
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•
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•
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•
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•
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•
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•
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•
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•
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•
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•
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•
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•
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•
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•
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